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Half controlled single bridge rectifier

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A half controlled single bridge rectifier is supplying an R-L load. It is operated at firing angle 'alpha'. Load current is continous. What is fraction of cycle that the freewheeling diode conducts?

    2. Relevant equations
    upload_2016-12-1_14-52-25.png
    3. The attempt at a solution
    Well I guess each diode will conduct for half the cycle like from 0 to pie and then from pie to 2 pie, the other diode will conduct.
    So the answer must be 1/2. But book answer is alpha / pie.
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2016 #2

    NascentOxygen

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    Staff: Mentor

    None of the 4 devices shown in your OP is a free-wheeling diode. Why is there no such diode shown in a question that apparently refers to it? :oldconfused:
     
  4. Dec 1, 2016 #3
  5. Dec 1, 2016 #4

    NascentOxygen

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    Staff: Mentor

    FWD in your attachment is a freewheeling diode.
     
  6. Dec 1, 2016 #5
    So it's a diode parallel to RL load with anode of diode towards up and cathode facing down.
    Thanks.
     
  7. Dec 1, 2016 #6

    NascentOxygen

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    Staff: Mentor

    If load current is continuous, then the diodes in the rectifier must share conduction equally and your original answer would have to be right.

    There is "freewheeling" operation due to the load inductance, otherwise load current could not be continuous. But I am not sure whether the 2 diodes that form half of the bridge are usually referred to as freewheeling diodes, though. I'll have to look at this more closely.

    It's the diode directly across the load that I think of as the classic freewheeling diode.
     
  8. Dec 1, 2016 #7
    I tried with ur concept of FWD and got this answer as:
    upload_2016-12-1_18-58-36.png
    upload_2016-12-1_18-59-33.png

    Thus i get the book answer.
    Book has also given another explanation. They have taken fwd as diode D1 and D2.
    They said that from
    1) from alpha to pie, current flows through T1, load, D2
    2) from pie to pie + alpha, current flows from T1 load, D1
    So D1 is in freewheeling mode only in second case. Duration is pie + alpha - pie which is alpha. This happens in each half cycle of sine wave, so twice in full cycle.
    Fraction is 2 * alpha divide by 2 * pie which gives alpha/pie.

    But i like first explanation of mine better where ur concept of FWD is used. Thanks..
     

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